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Re: [Phys-L] weighting in the wings ... damped harmonic oscillator ... bandwidth ... algebra ... bug hunting



Regarding:

I'm confused by some of the details in what David Bowman
wrote on 11/19/2016 08:44 PM, but overall the note was
helpful. It helped me clarify my thinking and my writing
on the following points:

You may wonder, why are we even talking about area under the
curve? If you take the classic Nyquist formula for the Johnson
noise, it is just

<|V|^2> = 4 kT R B [1]

If you slavishly apply the formula, all you need is the
bandwidth, not the gain or the area under the curve.

However, as usual, it pays to understand where the formula
comes from. I claim the real physics says:

<|V|^2> = ∫ 4 kT R |G|^2 df [2]

integrated over circular (not angular) frequency. Now in some
ideal world where the voltage gain G is unity within a passband
of width B and zero everywhere else, then equation [2] reduces
to [1]. However, in the other 99.999999% of the cases, you
actually have to do the integral. Figure out what's happening
at each frequency, and then add up all the contributions.

This is relevant to the RLC circuit, because the circuit does
not merely "select" the noise voltage components within the
bandwidth of the resonance; it also amplifies those components
by a factor of Q or so, just due to stuff sloshing around at
resonance. Even though it's a passive circuit, it still has
loads of gain.

Equation [1] will give the wrong answer for |V|^2, wrong by a
huge factor, on the order of Q^2.

The point of my previous note was that even after you figure out
that you need to look at the area under the curve, you can't
approximate it by height times FWHM, because there is a lot of
weight in the wings.

Here is some evidence that my calculation of the area under the
curve is correct, and my general way of looking at things is
correct.
Let's check the average /energy/ in the capacitor. It had darn
well better be ½ kT, independent of all circuit parameters.

This is calculated and discussed in a newly added subsection:
https://www.av8n.com/physics/rlc.htm#sec-thermo

There's also a new discussion of bandwidth, corner frequency,
and other frequency-like quantities at:
https://www.av8n.com/physics/rlc.htm#sec-freq-like

You may need to hit <reload> on your browser.

Since JD is confused about the details of my thinking in my previous post (and since I finally got some time to revisit the problem) I elaborate on my thinking and the details here.

There are a couple of definite integrals whose values are useful for this problem. I'll call them I_1 & I_2.

I_1 ≡ ∫(u=0 to u=∞){du/(1 + Q^2*(u - 1/u)^2)} = π/(2*Q)

and

I_2 ≡ ∫(u=0 to u=∞){du/((u/Q)^2 + (u^2 - 1)^2)} = π*Q/2

The former (I_1)appears in the calculation of the total power dissipated by the resistor in the LRC circuit in response to the thermal noise, and it also appears in the calculation of the average energy stored in the self-inductance. The latter integral (I_2) appears in the calculation of the average electrostatic energy stored in the capacitor.

Consider first the total (complex) series impedance of the circuit

Z_t = R + i*(ω*L - 1/(ω*C)) = R*(1 + i*Q*(f/f_0 - f_0/f)

where f is the (circular frequency), f_0 ≡ 1/(2*π*sqrt(L*C)), and Q ≡ sqrt(L/C)/R. The complex current amplitude I will be the noise voltage amplitude V divided by Z_t. The mean square current at frequency f is <I^2> = <V^2>/|Z_t|^2. Since the mean square noise voltage in some small frequency interval df is given by the Nyquist formula:

<V^2> = 4*k*T*R*df

this means the mean square current in that frequency interval is given by

<I^2> = 4*k*T*df/(R*(1 + Q^2*(f/f_0 - f_0/f)^2))

Since the power dissipated in any resistor is (I^2)*R this means the dissipated mean power in the frequency interval is

<P> = 4*k*T*df/(1 + Q^2*(f/f_0 - f_0/f)^2)

The total mean integrated power at all frequencies is

<P>_t = ∫(f=0 to f=∞){df*4*k*T/(1 + Q^2*(f/f_0 - f_0/f)^2)}

= 4*k*T*f_0*I_1 = 4*k*T*f_0*π/(2*Q) = kT*(2*π*f_0)/Q = k*T*ω_0/Q

If we work backward from this total integrated power and write it as

<P>_t = (<V^2>_t)/R

this gives us an effective expression for the total mean square voltage integrated over all frequencies as

<V^2>_t = kT*R*(2*π*f_0)/Q = 4*k*T*R*B

Where the integrated bandwidth, B in the Nyquist formula is

B = π*f_0/(2*Q) = f_0*I_1 .

If we look at the FWHM for the "gain" factor

1/(1 + Q^2*(f/f_0 - f_0/f)^2))

it is given by FWHM = Δf = f*0*sqrt(1 + 4*Q^2)/Q.

If we take the integrated bandwidth B & divide it by the FWHM Δf we get the ratio

B/Δf = π/(2*sqrt(1 + 4*Q^2))

as I claimed in my previous post.

Since the integrated mean square current is related to the total integrated power by <I^2>_t = <P>_t/R this means we can write that total integrated mean square current as

<I^2>_t = kT*(2*π*f_0)/(Q*R) .

If we wish to find the mean energy in the self-inductance we can save some work if we simply realize that since that energy is proportional to the total integrated mean square current over all frequencies. In particular

<E>_L = (L/2)*<I^2>_t = (L/2)*kT*(2*π*f_0)/(Q*R) =(L/R)*kT*π*f_0/Q .

If we remember that L/R = Q/(2*π*f_0) that means we can write

<E>_L = (Q/(2*π*f_0))*kT*π*f_0/Q = kT/2

as it should by the equipartition theorem.

If we want to find the energy in the capacitance we need the mean square voltage across the capacitor. We can find the voltage by multiplying the current amplitude by the impedance of the capacitor. This means the mean square voltage on the capacitor in some small frequency interval df is

<V^2>_C = |Z_C|^2*<I^2> = <I^2>/(ω*C)^2 .

Using the earlier above equation for <I^2> in the frequency interval we get


<V^2>_C = (4*k*T*df/(R*(1 + Q^2*(f/f_0 - f_0/f)^2)))/(2*π*f*C)^2

= 4*k*T*df/(R*(2*π*f*C)^2*(1 + Q^2*(f/f_0 - f_0/f)^2))

= 4*k*T*R*df/((f/(f_0*Q))^2*(1 + Q^2*(f/f_0 - f_0/f)^2))

= 4*k*T*R*df/((f/(f_0*Q))^2 + ((f/f_0)^2 - 1)^2) .

Next we note that the energy stored in the capacitor is (C/2)*V^2 so this means the mean energy in the capacitor in the frequency interval is

<E>_C = 2*k*T*R*C*df/((f/(f_0*Q))^2 + ((f/f_0)^2 - 1)^2)

= 2*k*T*(1/(2*π*f_0*Q))*df/((f/(f_0*Q))^2 + ((f/f_0)^2 - 1)^2)

= (k*T/(π*Q))*(1/f_0)*df/((f/(f_0*Q))^2 + ((f/f_0)^2 - 1)^2)

where we used the fact that R*C = 1/(2*π*f_0*Q). To get the total mean integrated energy on the capacitor we integrate this over all frequencies giving

<E>_C = ∫(f=0 to f=∞){(k*T/(π*Q))*(1/f_0)*df/((f/(f_0*Q))^2 + ((f/f_0)^2 - 1)^2)}

Factoring out the constants and letting u = f/f_0 be a new dimensionless integration variable gives

<E>_C = (k*T/(π*Q))*∫(u=0 to u=∞){du/((u/Q)^2 + (u^2 - 1)^2)

= (k*T/(π*Q))*I_2

= (k*T/(π*Q))*π*Q/2 = k*T/2

as it should by the equipartition theorem. In the last line here we used the value of the dimensionless integral I_2 = π*Q/2

So it looks like everything is normalized properly and the corresponding formulas all give the correct equipartition results.

David Bowman